In condensed matter physics, one often encounter a Hamiltonian of the form
$$\mathcal{H}=\sum_{\bf{k}}
\begin{pmatrix}a_{\bf{k}}^\dagger & a_{-\bf{k}}\end{pmatrix}
\begin{pmatrix}A_{\bf{k}} & B_{\bf{k}}\\B_{\bf{k}} & A_{\bf{k}}\end{pmatrix}
\begin{pmatrix}a_{\bf{k}} \\ a_{-\bf{k}}^\dagger\end{pmatrix}
,$$
where $a_{\bf{k}}$ is a bosonic operator. A Bogoliubov transformation
$$
\begin{pmatrix}a_{\bf{k}} \\ a_{-\bf{k}}^\dagger\end{pmatrix}=
\begin{pmatrix}\cosh\theta_{\bf{k}} & -\sinh\theta_{\bf{k}}\\-\sinh\theta_{\bf{k}} & \cosh\theta_{\bf{k}}\end{pmatrix}
\begin{pmatrix}\gamma_{\bf{k}} \\ \gamma_{-\bf{k}}^\dagger\end{pmatrix},
$$
with $$\tanh 2\theta_{\bf{k}}=\frac{B_{\bf{k}}}{A_{\bf{k}}}$$
is often used to diagonalized such a Hamiltonian. However, this seems to assume that
$|A_{\bf{k}}|> |B_{\bf{k}}|$. Is this true? If so, how else can the Hamiltonian be diagonalized?

where
$$\tag{5} \eta
~:=~\begin{pmatrix}1& 0\\0 & -1\end{pmatrix}$$
is the $1+1$ dimensional Minkowski metric. The Lie group $U(1,1)$ of Bogoliubov transformations is real and non-compact, and it is $4$-dimensional. In fact, one may prove that an element $U\in U(1,1)$ is of the form

Hence $|A|>|B|$ or $|A|=|B|$. In the latter case, $A^{\prime}=0$, so $N=0$, and hence $M=0$, and in particular $B=0$.

The proofs of (ii) and (iii) are left as exercises.

3) We will next argue that the case $A<|B|$ is not physically relevant, cf. Theorem II.

Theorem II:

(i) The Hamiltonian $H$ is positive definite if $A >|B|$.

(ii) The spectrum of $H$ is non-negative if $A\geq |B|$.

(iii) The spectrum of $H$ is unbounded from below if $A < |B|$.

A proof of Theorem II is left as an exercise. A proof at the classical level can be established by replacing the operators $a_1$ and $a_2$ with the corresponding classical complex variables, and then investigate the signature of the Hessian for the corresponding classical Hamiltonian function.

4) Finally, the Bogoliubov transformation (3) may be cast in a special relativistic language. One may view the matrix $M$, or equivalently $(A,B)$, as a point in $1+2$ dimensional Minkowski space with time coordinate $A\in \mathbb{R}$ and space coordinates $B\in \mathbb{C}\cong \mathbb{R}^2$. The invariant Minkowski length

$$\tag{14} \det(M)~=~A^2-|B|^2$$

is preserved under the action

$$\tag{15} \rho:U~\mapsto~ (U^{-1})^{\dagger}MU^{-1}$$

of the Lie group $U(1,1)$ of Bogoliubov transformations. Therefore $\rho$ is a Lie group homomorphism

$$\tag{16} \rho: \quad\to\quad O(1,2;\mathbb{R}) $$

into the $3$-dimensional Lorentz group $O(1,2;\mathbb{R})$. The condition $|A| >|B|$ ($|A| < |B|$) is the condition for being a time-like (space-like) vector, respectively. Intuitively, OP's observation concerning diagonalizability may be understood as the fact that one cannot turn a space-like vector into time-like vector by a Lorentz transformation.

For $A$ and $B$ real, using the parametrization given in the Wikipedia article on BT, we would end up with the condition $B[\cos(\theta_1-\theta_2)\cosh 2r+i\sin(\theta_1-\theta_2)]=A\sinh 2r$. This will lead to the same assumption on $A$ and $B$.
–
leongzFeb 7 '13 at 3:07

The Bogoliubov transformation is usually used to obtain excited states above a mean-field-type approximated ground states. Whenever the case $A\lt |B|$ arises, one formally obtains imaginary excitation energies, which means that the "excited states" actually drift away from the assumed ground state exponentially (in time), instead of displaying small amplitude harmonic oscillations, and imply an instability of the assumed ground state. This is a clear indication that the assumed ground state is not (even approximately) correct.
–
PapagenoJun 24 at 16:24